除法(Division)

最新推荐文章于 2020-07-30 15:56:13 发布

葡萄干是个程序员

最新推荐文章于 2020-07-30 15:56:13 发布

阅读量1.7k

点赞数

CC 4.0 BY-SA版权

分类专栏：暴力枚举文章标签： java UVa Division 除法简单枚举

本文链接：https://blog.youkuaiyun.com/qq_15096707/article/details/43951861

暴力枚举专栏收录该内容

7 篇文章

订阅专栏

本文介绍了一个编程挑战，任务是找出所有符合条件的五位数对，这些数对使用0到9的数字各一次，并且第一个数除以第二个数等于给定整数N。文章提供了Java实现代码及样例输入输出。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Division

Time Limit:3000MS

Memory Limit:Unknown

64bit IO Format:%lld & %llu

Submit Status

Description

Write a program that finds and displays all pairs of 5-digit numbers that between them use the digits 0 through 9 once each, such that the first number divided by the second is equal to an integer N, where $2\le N \le 79$ . That is,

abcde / fghij =N

where each letter represents a different digit. The first digit of one of the numerals is allowed to be zero.

Input

Each line of the input file consists of a valid integer N. An input of zero is to terminate the program.

Output

Your program have to display ALL qualifying pairs of numerals, sorted by increasing numerator (and, of course, denominator).

Your output should be in the following general form:

xxxxx / xxxxx =N

In case there are no pairs of numerals satisfying the condition, you must write ``There are no solutions for N.". Separate the output for two different values of N by a blank line.

Sample Input

61
62
0

Sample Output

There are no solutions for 61.

79546 / 01283 = 62
94736 / 01528 = 62

【分析】

枚举abcde，然后通过N算出fghij，最后判断是否所有数字不相同（统计0到9的个数，如果个数均为1，说明所有数字均不相同）。

用java语言编写程序，代码如下：

import java.io.BufferedInputStream;
import java.util.Scanner;

public class Main {
	public static void main(String[] args) {
		Scanner input = new Scanner(new BufferedInputStream(System.in));
		boolean first = true;
		while(input.hasNext()) {
			int n = input.nextInt();
			if(n == 0)
				break;
			
			if(first)
				first = !first;
			else
				System.out.println();
			
			boolean hasResult = false;
			for(int i = 123; i <= 98765; i++) {
				if(i % n == 0) {
					int m = i / n;
					if(count(i, m)) {
						hasResult = true;
						
						System.out.printf("%05d / %05d = " + n + "\n", i, m);
					}
				}
			}
			
			if(!hasResult)
				System.out.println("There are no solutions for " + n + ".");
		}
	}
	
	public static boolean count(int n, int m) {
		int[] count = new int[10];
		for(int i = 0; i < 5; i++) {
			count[n % 10]++;
			count[m % 10]++;
			
			if(count[n % 10] > 1 || count[m % 10] > 1)
				return false;
			
			n = n / 10;
			m = m / 10;
		}
		
		for(int i = 0; i < 10; i++) {
			if(count[i] != 1)
				return false;
		}
		return true;
	}
}